531. Lonely Pixel I (Medium)

Given a picture consisting of black and white pixels, find the number of black lonely pixels.

The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.

A black lonely pixel is character 'B' that located at a specific position where the same row and same column don't have any other black pixels.

Example:

Input: 
[['W', 'W', 'B'],
 ['W', 'B', 'W'],
 ['B', 'W', 'W']]

Output: 3
Explanation: All the three 'B's are black lonely pixels.

Note:

1. The range of width and height of the input 2D array is [1,500].

Solution:

$$O(nm)$$ Time, $$O(n+m)$$ Space

class Solution {
public:
    int findLonelyPixel(vector<vector<char>>& picture) {
        int m = picture.size(), n = picture[0].size(), cnt = 0;
        vector<int> row(m,0), col(n, 0);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (picture[i][j] == 'B') {
                    ++row[i]; ++col[j];
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (picture[i][j] == 'B' && row[i] == 1 && col[j] == 1) ++cnt;
            } 
        }
        return cnt;
    }
};

$$O(mn)$$ Time, $$O(n)$$ space

public class Solution {
    public int findLonelyPixel(char[][] picture) {
        if (picture == null || picture.length == 0 || picture[0].length == 0) return 0;

        int[] colArray = new int[picture[0].length];
        for (int i = 0; i < picture.length; i++) {
            for (int j = 0; j < picture[0].length; j++) {
                if (picture[i][j] == 'B') colArray[j]++;
            }
        }

        int ret = 0;
        for (int i = 0; i < picture.length; i++) {
            int count = 0, pos = 0;
            for (int j = 0; j < picture[0].length; j++) {
                if (picture[i][j] == 'B') {
                    count++;
                    pos = j;
                }
            }
            if (count == 1 && colArray[pos] == 1) ret++;
        }
        return ret;
    }
}

$$O(nm)$$ Time, $$O(1)$$ Space

public int findLonelyPixel(char[][] picture) {
    int n = picture.length, m = picture[0].length;

    int firstRowCount = 0;
    for (int i=0;i<n;i++) 
        for (int j=0;j<m;j++) 
            if (picture[i][j] == 'B') {   
                if (picture[0][j] < 'Y') picture[0][j]++;
                if (i == 0) firstRowCount++;
                else if (picture[i][0] < 'Y') picture[i][0]++;
            }

    int count = 0;
    for (int i=0;i<n;i++) 
        for (int j=0;j<m;j++) 
            if (picture[i][j] < 'W' && (picture[0][j] == 'C' || picture[0][j] == 'X')) { 
                if (i == 0) count += firstRowCount == 1 ? 1 : 0;
                else if (picture[i][0] == 'C' || picture[i][0] == 'X') count++;
            }

    return count;
}